OP Malhotra Class 11 Maths Solutions Chapter 19 Differentiation Exercise 19 (C)

S Chand Class 11 ICSE Maths Solutions Chapter 19 Differentiation Ex 19(C)

Question 1. Differentiate the following expression with respect to \(x\): \( (ax + b) (cx + d) \)
Answer:
Let \( y = (ax + b) (cx + d) \).
To find the derivative, we differentiate both sides of the equation with respect to \(x\). We will use the product rule for differentiation, which states that if \( y = uv \), then \( \frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx} \).
\( \frac { dy }{ dx } = (ax + b)\frac { d }{ dx }(cx + d) + (cx + d)\frac { d }{ dx }(ax + b) \)
\( \implies \frac { dy }{ dx } = (ax + b)(c \cdot 1 + 0) + (cx + d)(a \cdot 1 + 0) \)
\( \implies \frac { dy }{ dx } = c(ax + b) + a(cx + d) \)
\( \implies \frac { dy }{ dx } = acx + bc + acx + ad \)
\( \implies \frac { dy }{ dx } = 2acx + bc + ad \)
In simple words: To differentiate two terms multiplied together, we use the product rule. This rule says you differentiate one term while keeping the other the same, and then add that to differentiating the second term while keeping the first one the same. After applying this, we combine similar terms.

🎯 Exam Tip: Remember the product rule formula: \( \frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx} \). Be careful with constant terms and variables during differentiation.

Question 2. Differentiate the following expression with respect to \(x\): \( (x^{100} + 2x^{50} – 3) (7x^8 + 20x + 5) \)
Answer:
Let \( y = (x^{100} + 2x^{50} – 3) (7x^8 + 20x + 5) \).
We differentiate both sides with respect to \(x\), using the product rule. This rule is very useful when two functions of \(x\) are multiplied together.
\( \frac { dy }{ dx } = (x^{100} + 2x^{50} – 3) \frac { d }{ dx } (7x^8 + 20x + 5) + (7x^8 + 20x + 5) \frac { d }{ dx } (x^{100} + 2x^{50} – 3) \)
\( \implies \frac { dy }{ dx } = (x^{100} + 2x^{50} – 3) (7 \cdot 8x^{8-1} + 20 \cdot 1 + 0) + (7x^8 + 20x + 5) (100x^{100-1} + 2 \cdot 50x^{50-1} - 0) \)
\( \implies \frac { dy }{ dx } = (x^{100} + 2x^{50} – 3) (56x^7 + 20) + (7x^8 + 20x + 5) (100x^{99} + 100x^{49}) \)
In simple words: This problem also uses the product rule. You differentiate the first part and multiply by the second, then add that to differentiating the second part and multiplying by the first. Remember to apply the power rule \( \frac{d}{dx}(x^n) = nx^{n-1} \) for each term.

🎯 Exam Tip: When differentiating polynomials, treat each term separately. Constants differentiate to zero, and the power rule is key for terms like \(x^n\).

Question 3. Differentiate the following expression with respect to \(x\): \( x (2x – 1)(x + 2) \)
Answer:
Let \( y = x (2x – 1) (x + 2) \).
We differentiate both sides with respect to \(x\). For a product of three functions \(uvw\), the derivative is \( \frac{d}{dx}(uvw) = uv\frac{dw}{dx} + uw\frac{dv}{dx} + vw\frac{du}{dx} \). This expands the product rule for more terms.
\( \frac { dy }{ dx } = x(2x – 1)\frac { d }{ dx }(x + 2) + x (x + 2) \frac { d }{ dx }(2x – 1) + (2x – 1)(x + 2)\frac { d }{ dx }x \)
\( \implies \frac { dy }{ dx } = x(2x – 1)(1 + 0) + x (x + 2) (2 - 0) + (2x – 1)(x + 2)(1) \)
\( \implies \frac { dy }{ dx } = x(2x – 1) + 2x(x + 2) + (2x – 1)(x + 2) \)
In simple words: When you have three things multiplied, you take turns differentiating each one while keeping the other two the same, and then add up these three results. This is an extension of the basic product rule.

🎯 Exam Tip: For products of three or more functions, you can group them (e.g., \(u(vw)\)) and apply the product rule iteratively, or use the extended formula directly.

Question 4. Differentiate the following expression with respect to \(x\): \( (x – 2) (x + 3) (2x + 5) \)
Answer:
Let \( y = (x – 2) (x + 3) (2x + 5) \).
We differentiate both sides with respect to \(x\), using the extended product rule for three functions: \( \frac{d}{dx}(uvw) = uv\frac{dw}{dx} + uw\frac{dv}{dx} + vw\frac{du}{dx} \). This helps to differentiate each part systematically.
\( \frac { dy }{ dx } = (x – 2) (x + 3) \frac { d }{ dx } (2x + 5) + (x – 2) (2x + 5) \frac { d }{ dx }(x + 3) + (x + 3) (2x + 5) \frac { d }{ dx }(x – 2) \)
\( \implies \frac { dy }{ dx } = (x – 2) (x + 3) (2 \cdot 1 + 0) + (x – 2) (2x + 5) (1 + 0) + (x + 3) (2x + 5) (1 – 0) \)
\( \implies \frac { dy }{ dx } = 2(x – 2) (x + 3) + (x – 2) (2x + 5) + (x + 3) (2x + 5) \)
In simple words: This is another case of multiplying three terms. To differentiate, we differentiate each term one by one, multiplying it by the other two terms kept unchanged, and then sum these results. Remember that the derivative of \( (x \pm c) \) is just 1.

🎯 Exam Tip: Differentiating a linear term like \( (ax+b) \) always results in just \(a\). For \( (x-2) \), \( (x+3) \), \( (2x+5) \), their derivatives are \(1, 1, 2\) respectively.

Question 5. Differentiate the following expression with respect to \(x\): \( y = \frac{2x+5}{3x-2} \)
Answer:
Let \( y = \frac{2x+5}{3x-2} \).
We differentiate both sides with respect to \(x\). Since this is a fraction, we apply the quotient rule. The quotient rule states that if \( y = \frac{u}{v} \), then \( \frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \), provided \(v \neq 0\). This rule helps us find the derivative of rational functions.
\( \frac{dy}{dx} = \frac{(3x-2)\frac{d}{dx}(2x+5) - (2x+5)\frac{d}{dx}(3x-2)}{(3x-2)^2} \)
\( \implies \frac{dy}{dx} = \frac{(3x-2)(2+0) - (2x+5)(3-0)}{(3x-2)^2} \)
\( \implies \frac{dy}{dx} = \frac{2(3x-2) - 3(2x+5)}{(3x-2)^2} \)
\( \implies \frac{dy}{dx} = \frac{6x-4 - 6x-15}{(3x-2)^2} \)
\( \implies \frac{dy}{dx} = \frac{-19}{(3x-2)^2} \)
In simple words: For a fraction, we use the quotient rule. This rule involves multiplying the bottom term by the derivative of the top term, then subtracting the top term multiplied by the derivative of the bottom term. All of this is then divided by the square of the bottom term.

🎯 Exam Tip: When using the quotient rule, be very careful with the order of subtraction in the numerator. It's always \(v\frac{du}{dx} - u\frac{dv}{dx}\), not the other way around.

Question 6. Differentiate the following expression with respect to \(x\): \( y = \frac{x^2-3}{x+4} \)
Answer:
Let \( y = \frac{x^2-3}{x+4} \).
To find the derivative, we differentiate both sides with respect to \(x\). Since this expression is a fraction, we must use the quotient rule for differentiation, which helps find the derivative of a ratio of two functions.
\( \frac{dy}{dx} = \frac{(x+4)\frac{d}{dx}(x^2-3) - (x^2-3)\frac{d}{dx}(x+4)}{(x+4)^2} \)
\( \implies \frac{dy}{dx} = \frac{(x+4)(2x-0) - (x^2-3)(1+0)}{(x+4)^2} \)
\( \implies \frac{dy}{dx} = \frac{2x(x+4) - (x^2-3)}{(x+4)^2} \)
\( \implies \frac{dy}{dx} = \frac{2x^2+8x - x^2+3}{(x+4)^2} \)
\( \implies \frac{dy}{dx} = \frac{x^2+8x+3}{(x+4)^2} \)
In simple words: This problem requires the quotient rule for fractions. You take the bottom function times the derivative of the top, minus the top function times the derivative of the bottom, all divided by the bottom function squared.

🎯 Exam Tip: Always expand the numerator carefully and combine like terms after applying the quotient rule to simplify the expression fully.

Question 7. Differentiate the following expression with respect to \(x\): \( y = \frac{2x-3}{3x+4} \)
Answer:
Let \( y = \frac{2x-3}{3x+4} \).
We differentiate both sides with respect to \(x\). Since we have a fraction, we apply the quotient rule for derivatives, which helps in calculating the rate of change of one function divided by another.
\( \frac{dy}{dx} = \frac{(3x+4)\frac{d}{dx}(2x-3) - (2x-3)\frac{d}{dx}(3x+4)}{(3x+4)^2} \)
\( \implies \frac{dy}{dx} = \frac{(3x+4)(2 \cdot 1-0) - (2x-3)(3 \cdot 1+0)}{(3x+4)^2} \)
\( \implies \frac{dy}{dx} = \frac{2(3x+4) - 3(2x-3)}{(3x+4)^2} \)
\( \implies \frac{dy}{dx} = \frac{6x+8 - 6x+9}{(3x+4)^2} \)
\( \implies \frac{dy}{dx} = \frac{17}{(3x+4)^2} \)
In simple words: This is another example of using the quotient rule. The key is to correctly apply the formula by multiplying the denominator by the derivative of the numerator and subtracting the numerator times the derivative of the denominator, then dividing by the denominator squared.

🎯 Exam Tip: Pay close attention to signs, especially when distributing the negative sign in the numerator of the quotient rule. A small error can lead to a wrong final answer.

Question 8. Differentiate the following expression with respect to \(x\): \( y = \frac{x^5-x+2}{x^3+7} \)
Answer:
Let \( y = \frac{x^5-x+2}{x^3+7} \).
We differentiate both sides with respect to \(x\). Given the form of the expression, we use the quotient rule to find its derivative. This rule is essential for functions that are ratios of two other functions.
\( \frac{dy}{dx} = \frac{(x^3+7)\frac{d}{dx}(x^5-x+2) - (x^5-x+2)\frac{d}{dx}(x^3+7)}{(x^3+7)^2} \)
\( \implies \frac{dy}{dx} = \frac{(x^3+7)(5x^4-1+0) - (x^5-x+2)(3x^2+0)}{(x^3+7)^2} \)
\( \implies \frac{dy}{dx} = \frac{(x^3+7)(5x^4-1) - (x^5-x+2)(3x^2)}{(x^3+7)^2} \)
\( \implies \frac{dy}{dx} = \frac{(5x^7-x^3+35x^4-7) - (3x^7-3x^3+6x^2)}{(x^3+7)^2} \)
\( \implies \frac{dy}{dx} = \frac{5x^7-x^3+35x^4-7 - 3x^7+3x^3-6x^2}{(x^3+7)^2} \)
\( \implies \frac{dy}{dx} = \frac{2x^7+2x^3+35x^4-6x^2-7}{(x^3+7)^2} \)
In simple words: When you need to differentiate a fraction, always use the quotient rule. It's about combining the derivative of the top and bottom parts correctly, always dividing by the square of the bottom part. Make sure to multiply out terms and simplify carefully.

🎯 Exam Tip: Be meticulous when expanding products and combining terms in the numerator. A single mistake in multiplication or signs can alter the entire derivative.

Question 9. Differentiate the following expression with respect to \(t\): \( s = t^2 (t + 1)^{-1} \)
Answer:
Let \( s = t^2 (t + 1)^{-1} \). We can also write this as \( s = \frac{t^2}{t+1} \).
We differentiate both sides with respect to \(t\). Since the expression is in a fractional form, the quotient rule is the most suitable method for finding the derivative of this function.
\( \frac{ds}{dt} = \frac{(t+1)\frac{d}{dt}(t^2) - t^2\frac{d}{dt}(t+1)}{(t+1)^2} \)
\( \implies \frac{ds}{dt} = \frac{(t+1)(2t) - t^2(1)}{(t+1)^2} \)
\( \implies \frac{ds}{dt} = \frac{2t^2+2t - t^2}{(t+1)^2} \)
\( \implies \frac{ds}{dt} = \frac{t^2+2t}{(t+1)^2} \)
\( \implies \frac{ds}{dt} = \frac{t(t+2)}{(t+1)^2} \)
In simple words: This problem involves a term with a negative exponent, which means it's a fraction. So, we use the quotient rule. Differentiate with respect to \(t\), keeping track of all the \(t\) terms.

🎯 Exam Tip: Expressions with negative exponents can often be rewritten as fractions, making the application of the quotient rule more straightforward. Always simplify the final expression by factoring if possible.

Question 10. Differentiate the following expression with respect to \(u\): \( z = \frac{u}{u^2+1} \)
Answer:
Let \( z = \frac{u}{u^2+1} \).
We differentiate both sides with respect to \(u\). As this is a fractional expression, we apply the quotient rule, which helps us calculate the rate of change of a rational function. The rule is \( \frac{d}{du}\left(\frac{N}{D}\right) = \frac{D\frac{dN}{du} - N\frac{dD}{du}}{D^2} \).
\( \frac{dz}{du} = \frac{(u^2+1)\frac{d}{du}(u) - u\frac{d}{du}(u^2+1)}{(u^2+1)^2} \)
\( \implies \frac{dz}{du} = \frac{(u^2+1)(1) - u(2u)}{(u^2+1)^2} \)
\( \implies \frac{dz}{du} = \frac{u^2+1 - 2u^2}{(u^2+1)^2} \)
\( \implies \frac{dz}{du} = \frac{1-u^2}{(u^2+1)^2} \)
In simple words: For a function that is a fraction, use the quotient rule. This means the bottom part times the derivative of the top, minus the top part times the derivative of the bottom, all divided by the square of the bottom part.

🎯 Exam Tip: When differentiating with respect to a variable (e.g., \(u\)), treat other symbols as constants if they are not the variable of differentiation. Here, the derivative of \(u\) with respect to \(u\) is 1.

Question 11. Differentiate the following expression with respect to \(x\): \( y = \frac{x^2+2x+5}{x^3+2x+4} \)
Answer:
Let \( y = \frac{x^2+2x+5}{x^3+2x+4} \).
We differentiate both sides with respect to \(x\). Since this expression is a fraction of two functions, we apply the quotient rule to find its derivative. This rule helps us accurately determine the rate of change for such complex rational expressions.
\( \frac{dy}{dx} = \frac{(x^3+2x+4)\frac{d}{dx}(x^2+2x+5) - (x^2+2x+5)\frac{d}{dx}(x^3+2x+4)}{(x^3+2x+4)^2} \)
\( \implies \frac{dy}{dx} = \frac{(x^3+2x+4)(2x+2) - (x^2+2x+5)(3x^2+2)}{(x^3+2x+4)^2} \)
\( \implies \frac{dy}{dx} = \frac{(2x^4+2x^3+4x^2+4x+8x+8) - (3x^4+2x^2+6x^3+4x+15x^2+10)}{(x^3+2x+4)^2} \)
\( \implies \frac{dy}{dx} = \frac{2x^4+2x^3+4x^2+12x+8 - 3x^4-6x^3-17x^2-4x-10}{(x^3+2x+4)^2} \)
\( \implies \frac{dy}{dx} = \frac{-x^4-4x^3-13x^2+8x-2}{(x^3+2x+4)^2} \)
In simple words: When you differentiate a fraction, use the quotient rule. This means: (bottom times derivative of top) minus (top times derivative of bottom), all divided by (bottom squared). Be very careful when expanding and combining all the terms in the numerator.

🎯 Exam Tip: Algebraic simplification after applying the quotient rule can be lengthy. Double-check each multiplication and subtraction step to avoid errors. Group terms by power of \(x\) to ensure all like terms are combined correctly.

Question 12. Differentiate the following expression with respect to \(x\): \( f(x) = \frac{x^3+2x}{x^2+4} \)
Answer:
Let \( f(x) = \frac{x^3+2x}{x^2+4} \).
To find the derivative \( f'(x) \), we differentiate with respect to \(x\). As this is a rational function, we apply the quotient rule. The quotient rule is essential for correctly differentiating functions that are expressed as one function divided by another.
\( f'(x) = \frac{(x^2+4)\frac{d}{dx}(x^3+2x) - (x^3+2x)\frac{d}{dx}(x^2+4)}{(x^2+4)^2} \)
\( \implies f'(x) = \frac{(x^2+4)(3x^2+2) - (x^3+2x)(2x)}{(x^2+4)^2} \)
\( \implies f'(x) = \frac{(3x^4+2x^2+12x^2+8) - (2x^4+4x^2)}{(x^2+4)^2} \)
\( \implies f'(x) = \frac{3x^4+14x^2+8 - 2x^4-4x^2}{(x^2+4)^2} \)
\( \implies f'(x) = \frac{x^4+10x^2+8}{(x^2+4)^2} \)
In simple words: When you see a fraction with \(x\) terms in both the top and bottom, use the quotient rule to find the derivative. After applying the rule, carefully multiply and combine similar terms to get the simplest answer.

🎯 Exam Tip: When calculating \(f'(x)\), ensure that you fully expand and simplify the numerator. Leaving it unsimplified might result in loss of marks.

Question 13. If \( f(x) = \frac{x+2}{x-2} \) for all \(x \neq 2\), find \( f' ( – 2) \).
Answer:
Given \( f(x) = \frac{x+2}{x-2} \).
First, we differentiate \(f(x)\) with respect to \(x\) using the quotient rule. This rule is used because \(f(x)\) is a ratio of two functions of \(x\).
\( f'(x) = \frac{(x-2)\frac{d}{dx}(x+2) - (x+2)\frac{d}{dx}(x-2)}{(x-2)^2} \)
\( \implies f'(x) = \frac{(x-2)(1) - (x+2)(1)}{(x-2)^2} \)
\( \implies f'(x) = \frac{x-2 - x-2}{(x-2)^2} \)
\( \implies f'(x) = \frac{-4}{(x-2)^2} \)
Now, to find \( f'(-2) \), we substitute \(x = -2\) into the derivative.
\( f'(-2) = \frac{-4}{(-2-2)^2} \)
\( \implies f'(-2) = \frac{-4}{(-4)^2} \)
\( \implies f'(-2) = \frac{-4}{16} \)
\( \implies f'(-2) = -\frac{1}{4} \)
In simple words: First, find the derivative of the given function using the quotient rule. After you have the derivative, replace all the \(x\) values with \(-2\) to find the specific value of the derivative at that point.

🎯 Exam Tip: Always calculate the general derivative \(f'(x)\) first, and then substitute the specific value of \(x\). Do not substitute \(x\) before differentiating, as that would make the function a constant and its derivative zero.

Question 14. Differentiate \( \frac{x+2}{x^2-3} \) and find the value of the derivative at \( x = 0 \).
Answer:
Given \( f(x) = \frac{x+2}{x^2-3} \).
First, we differentiate \(f(x)\) with respect to \(x\) using the quotient rule. This rule is necessary when differentiating a function that is a fraction of two other functions.
\( f'(x) = \frac{(x^2-3)\frac{d}{dx}(x+2) - (x+2)\frac{d}{dx}(x^2-3)}{(x^2-3)^2} \)
\( \implies f'(x) = \frac{(x^2-3)(1) - (x+2)(2x)}{(x^2-3)^2} \)
\( \implies f'(x) = \frac{x^2-3 - (2x^2+4x)}{(x^2-3)^2} \)
\( \implies f'(x) = \frac{x^2-3 - 2x^2-4x}{(x^2-3)^2} \)
\( \implies f'(x) = \frac{-x^2-4x-3}{(x^2-3)^2} \)
Now, we find the value of the derivative at \(x = 0\) by substituting \(x=0\) into \(f'(x)\).
\( f'(0) = \frac{-(0)^2-4(0)-3}{((0)^2-3)^2} \)
\( \implies f'(0) = \frac{-3}{(-3)^2} \)
\( \implies f'(0) = \frac{-3}{9} \)
\( \implies f'(0) = -\frac{1}{3} \)
In simple words: First, find the derivative of the fraction using the quotient rule. Then, put \(x=0\) into the derivative formula you found to get the exact value of the derivative at that point.

🎯 Exam Tip: Remember that \(x^2-3\) in the denominator squared will become \((-3)^2=9\) when \(x=0\), not zero. Be careful with calculations involving zero and negative numbers.

Question 15. If \( y = \frac{x}{x+a} \), prove that \( x\frac{dy}{dx} = y (1 – y) \).
Answer:
Given \( y = \frac{x}{x+a} \).
First, we find \( \frac{dy}{dx} \) by differentiating \(y\) with respect to \(x\) using the quotient rule. This derivative is a key step in verifying the given proof.
\( \frac{dy}{dx} = \frac{(x+a)\frac{d}{dx}(x) - x\frac{d}{dx}(x+a)}{(x+a)^2} \)
\( \implies \frac{dy}{dx} = \frac{(x+a)(1) - x(1)}{(x+a)^2} \)
\( \implies \frac{dy}{dx} = \frac{x+a - x}{(x+a)^2} \)
\( \implies \frac{dy}{dx} = \frac{a}{(x+a)^2} \) (Equation 1)
Now, let's find the expression for \( y(1-y) \). We know \( y = \frac{x}{x+a} \).
\( 1-y = 1 - \frac{x}{x+a} = \frac{x+a-x}{x+a} = \frac{a}{x+a} \)
\( \implies y(1-y) = \frac{x}{x+a} \cdot \frac{a}{x+a} \)
\( \implies y(1-y) = \frac{ax}{(x+a)^2} \) (Equation 2)
Next, let's calculate \( x\frac{dy}{dx} \) using the derivative we found in Equation 1.
\( x\frac{dy}{dx} = x \cdot \frac{a}{(x+a)^2} \)
\( \implies x\frac{dy}{dx} = \frac{ax}{(x+a)^2} \) (Equation 3)
Comparing Equation 2 and Equation 3, we see that \( x\frac{dy}{dx} = y(1-y) \). Thus, the statement is proven.
In simple words: To prove this, first find the derivative of \(y\). Then, calculate both sides of the equation separately: \( x \) times the derivative, and \( y \) times \( (1-y) \). If both sides simplify to the same expression, the proof is complete.

🎯 Exam Tip: For proof questions, always show both sides of the equation are equal by transforming them independently or substituting one into the other. Clearly label intermediate results (like equations 1 and 2) for clarity.

Question 16. If \( x \sqrt{1+y}+y \sqrt{1+x}=0 \), prove that \( \frac{dy}{dx} = -\frac{1}{(1+x)^2} \).
Answer:
Given \( x \sqrt{1+y}+y \sqrt{1+x}=0 \).
First, we rearrange the equation to isolate terms, which simplifies the process of implicit differentiation. Since \(x \sqrt{1+y} = -y \sqrt{1+x}\), we can square both sides to remove the square roots.
\( x \sqrt{1+y} = -y \sqrt{1+x} \)
Squaring both sides:
\( (x \sqrt{1+y})^2 = (-y \sqrt{1+x})^2 \)
\( \implies x^2(1+y) = y^2(1+x) \)
\( \implies x^2+x^2y = y^2+y^2x \)
Rearrange the terms to group \(x\) and \(y\):
\( x^2-y^2 + x^2y-y^2x = 0 \)
Factor the terms:
\( (x-y)(x+y) + xy(x-y) = 0 \)
\( \implies (x-y)(x+y+xy) = 0 \)
From the original equation \( x \sqrt{1+y}+y \sqrt{1+x}=0 \), if \( x=y \), then \( x \sqrt{1+x}+x \sqrt{1+x}=0 \implies 2x \sqrt{1+x}=0 \). This implies \(x=0\) or \(1+x=0\). If \(x=0\), then \(y=0\), which makes \(0=0\). If \(1+x=0\), then \(x=-1\), which makes \(y=-1\), which is also \(0=0\). However, if \(x=y\), the solution becomes trivial or leads to specific points. The condition \( x \neq y \) is often implied in such problems to avoid such trivial cases, making \(x-y \neq 0\). Therefore, we can divide by \( (x-y) \).
So, \( x+y+xy = 0 \)
Now, we need to solve for \(y\) to find an explicit function, or we can use implicit differentiation.
\( y+xy = -x \)
\( y(1+x) = -x \)
\( y = \frac{-x}{1+x} \)
Now, we differentiate \(y\) with respect to \(x\) using the quotient rule.
\( \frac{dy}{dx} = \frac{(1+x)\frac{d}{dx}(-x) - (-x)\frac{d}{dx}(1+x)}{(1+x)^2} \)
\( \implies \frac{dy}{dx} = \frac{(1+x)(-1) - (-x)(1)}{(1+x)^2} \)
\( \implies \frac{dy}{dx} = \frac{-1-x + x}{(1+x)^2} \)
\( \implies \frac{dy}{dx} = \frac{-1}{(1+x)^2} \)
Thus, the proof is complete.
In simple words: This problem asks us to prove a derivative. First, rearrange the original equation by squaring both sides to remove square roots. Then, simplify the equation and solve for \(y\). Finally, differentiate \(y\) using the quotient rule. The result should match what we needed to prove.

🎯 Exam Tip: For implicit differentiation problems, squaring both sides can often simplify the expression before differentiation. Always try to isolate \(y\) if possible to simplify the differentiation process to an explicit one.

Question 17. Given that \( y = \sqrt{\frac{1-x}{1+x}} \), show that \( (1 – x^2)\frac{dy}{dx} + y = 0 \).
Answer:
Given \( y = \sqrt{\frac{1-x}{1+x}} \).
First, we differentiate \(y\) with respect to \(x\). We can rewrite \(y\) as \( y = \left(\frac{1-x}{1+x}\right)^{1/2} \).
We will use the chain rule and the quotient rule. The chain rule states \( \frac{d}{dx}f(g(x)) = f'(g(x))g'(x) \).
\( \frac{dy}{dx} = \frac{1}{2} \left(\frac{1-x}{1+x}\right)^{\frac{1}{2}-1} \frac{d}{dx}\left(\frac{1-x}{1+x}\right) \)
\( \implies \frac{dy}{dx} = \frac{1}{2} \left(\frac{1-x}{1+x}\right)^{-\frac{1}{2}} \frac{(1+x)\frac{d}{dx}(1-x) - (1-x)\frac{d}{dx}(1+x)}{(1+x)^2} \)
\( \implies \frac{dy}{dx} = \frac{1}{2} \left(\frac{1+x}{1-x}\right)^{\frac{1}{2}} \frac{(1+x)(-1) - (1-x)(1)}{(1+x)^2} \)
\( \implies \frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{1+x}{1-x}} \frac{-1-x-1+x}{(1+x)^2} \)
\( \implies \frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{1+x}{1-x}} \frac{-2}{(1+x)^2} \)
\( \implies \frac{dy}{dx} = -\sqrt{\frac{1+x}{1-x}} \frac{1}{(1+x)^2} \)
We can simplify the term \( \frac{1}{(1+x)^2} \) as \( \frac{1}{(1+x)\sqrt{1+x}\sqrt{1+x}} \).
\( \implies \frac{dy}{dx} = -\frac{\sqrt{1+x}}{\sqrt{1-x}} \frac{1}{(1+x)\sqrt{1+x}\sqrt{1+x}} \)
\( \implies \frac{dy}{dx} = -\frac{1}{\sqrt{1-x}\sqrt{1+x}(1+x)} \)
We can also write \( \sqrt{1-x}\sqrt{1+x} = \sqrt{1-x^2} \).
\( \implies \frac{dy}{dx} = -\frac{1}{(1+x)\sqrt{1-x^2}} \)
Now, let's substitute this and \(y\) back into the expression we need to prove: \( (1 – x^2)\frac{dy}{dx} + y = 0 \).
\( (1-x^2) \left( -\frac{1}{(1+x)\sqrt{1-x^2}} \right) + \sqrt{\frac{1-x}{1+x}} \)
We know \( (1-x^2) = \sqrt{1-x^2}\sqrt{1-x^2} \).
\( \implies \frac{\sqrt{1-x^2}\sqrt{1-x^2}}{1} \left( -\frac{1}{(1+x)\sqrt{1-x^2}} \right) + \sqrt{\frac{1-x}{1+x}} \)
\( \implies -\frac{\sqrt{1-x^2}}{1+x} + \sqrt{\frac{1-x}{1+x}} \)
\( \implies -\frac{\sqrt{(1-x)(1+x)}}{1+x} + \sqrt{\frac{1-x}{1+x}} \)
\( \implies -\frac{\sqrt{1-x}\sqrt{1+x}}{\sqrt{1+x}\sqrt{1+x}} + \sqrt{\frac{1-x}{1+x}} \)
\( \implies -\frac{\sqrt{1-x}}{\sqrt{1+x}} + \sqrt{\frac{1-x}{1+x}} \)
\( \implies -\sqrt{\frac{1-x}{1+x}} + \sqrt{\frac{1-x}{1+x}} \)
\( \implies 0 \)
Thus, \( (1 – x^2)\frac{dy}{dx} + y = 0 \) is proven.
In simple words: To prove this, first find the derivative of \(y\) using the chain rule and quotient rule. Then, substitute this derivative and the original \(y\) into the given equation \( (1 – x^2)\frac{dy}{dx} + y \). After careful simplification, both sides should become zero, completing the proof.

🎯 Exam Tip: When simplifying square root terms like \( \sqrt{1-x^2} \), remember that \( 1-x^2 = (1-x)(1+x) \). This factorization is often crucial for cancelling terms and simplifying expressions in differentiation proofs.

Question 18. Given that \( y = (3x – 1)^2 + (2x – 1)^3 \), find \( \frac{dy}{dx} \) and the points on the curve for which \( \frac{dy}{dx} = 0 \).
Answer:
Given \( y = (3x – 1)^2 + (2x – 1)^3 \).
First, we find \( \frac{dy}{dx} \) by differentiating \(y\) with respect to \(x\) using the chain rule. The chain rule is essential for differentiating composite functions like \((ax+b)^n\).
\( \frac{dy}{dx} = \frac{d}{dx}(3x-1)^2 + \frac{d}{dx}(2x-1)^3 \)
\( \implies \frac{dy}{dx} = 2(3x-1)^{2-1} \cdot \frac{d}{dx}(3x-1) + 3(2x-1)^{3-1} \cdot \frac{d}{dx}(2x-1) \)
\( \implies \frac{dy}{dx} = 2(3x-1)(3) + 3(2x-1)^2(2) \)
\( \implies \frac{dy}{dx} = 6(3x-1) + 6(2x-1)^2 \)
\( \implies \frac{dy}{dx} = 6[(3x-1) + (2x-1)^2] \)
\( \implies \frac{dy}{dx} = 6[3x-1 + (4x^2-4x+1)] \)
\( \implies \frac{dy}{dx} = 6[4x^2-x] \)
\( \implies \frac{dy}{dx} = 6x(4x-1) \)
Next, we find the points on the curve where \( \frac{dy}{dx} = 0 \).
\( 6x(4x-1) = 0 \)
This gives us two possible values for \(x\):
\( 6x = 0 \implies x = 0 \)
\( 4x-1 = 0 \implies 4x = 1 \implies x = \frac{1}{4} \)
Now, we find the corresponding \(y\) values for these \(x\) values using the original equation \( y = (3x – 1)^2 + (2x – 1)^3 \).
When \( x = 0 \):
\( y = (3(0)-1)^2 + (2(0)-1)^3 \)
\( \implies y = (-1)^2 + (-1)^3 \)
\( \implies y = 1 - 1 = 0 \)
So, one point is \( (0, 0) \).
When \( x = \frac{1}{4} \):
\( y = \left(3\left(\frac{1}{4}\right)-1\right)^2 + \left(2\left(\frac{1}{4}\right)-1\right)^3 \)
\( \implies y = \left(\frac{3}{4}-1\right)^2 + \left(\frac{1}{2}-1\right)^3 \)
\( \implies y = \left(-\frac{1}{4}\right)^2 + \left(-\frac{1}{2}\right)^3 \)
\( \implies y = \frac{1}{16} - \frac{1}{8} \)
\( \implies y = \frac{1-2}{16} = -\frac{1}{16} \)
So, the other point is \( \left(\frac{1}{4}, -\frac{1}{16}\right) \).
Therefore, the required points on the curve where \( \frac{dy}{dx} = 0 \) are \( (0, 0) \) and \( \left(\frac{1}{4}, -\frac{1}{16}\right) \).
In simple words: First, find the derivative of the function using the chain rule. Then, set this derivative equal to zero to find the \(x\)-values where the slope is flat. Finally, plug these \(x\)-values back into the original equation to find the matching \(y\)-values, giving you the points on the curve.

🎯 Exam Tip: Remember to use the chain rule for terms like \( (ax+b)^n \), which means \( n(ax+b)^{n-1} \cdot a \). When finding points where \( \frac{dy}{dx}=0 \), always provide both the \(x\) and \(y\) coordinates.

Question 19.
(i) If \( y = \frac{x-1}{2x^2-7x+5} \), find \( \frac{dy}{dx} \) at \( x = 2 \).
(ii) If \( y = \frac{x^2+3}{x^3+2x} \), find \( \frac{dy}{dx} \) at \( x = 1 \).
Answer:
(i) Given \( y = \frac{x-1}{2x^2-7x+5} \).
First, we factor the denominator: \( 2x^2-7x+5 = (2x-5)(x-1) \).
So, \( y = \frac{x-1}{(2x-5)(x-1)} \). For \( x \neq 1 \) and \( x \neq \frac{5}{2} \), we can simplify this to:
\( y = \frac{1}{2x-5} = (2x-5)^{-1} \).
Now, we differentiate \(y\) with respect to \(x\) using the chain rule.
\( \frac{dy}{dx} = -1(2x-5)^{-1-1} \cdot \frac{d}{dx}(2x-5) \)
\( \implies \frac{dy}{dx} = -1(2x-5)^{-2} \cdot 2 \)
\( \implies \frac{dy}{dx} = -\frac{2}{(2x-5)^2} \)
Now, we find the value of \( \frac{dy}{dx} \) at \( x = 2 \).
\( \left(\frac{dy}{dx}\right)_{x=2} = -\frac{2}{(2(2)-5)^2} \)
\( \implies \left(\frac{dy}{dx}\right)_{x=2} = -\frac{2}{(4-5)^2} \)
\( \implies \left(\frac{dy}{dx}\right)_{x=2} = -\frac{2}{(-1)^2} \)
\( \implies \left(\frac{dy}{dx}\right)_{x=2} = -\frac{2}{1} = -2 \)

(ii) Given \( y = \frac{x^2+3}{x^3+2x} \).
We differentiate \(y\) with respect to \(x\) using the quotient rule.
\( \frac{dy}{dx} = \frac{(x^3+2x)\frac{d}{dx}(x^2+3) - (x^2+3)\frac{d}{dx}(x^3+2x)}{(x^3+2x)^2} \)
\( \implies \frac{dy}{dx} = \frac{(x^3+2x)(2x) - (x^2+3)(3x^2+2)}{(x^3+2x)^2} \)
\( \implies \frac{dy}{dx} = \frac{2x^4+4x^2 - (3x^4+2x^2+9x^2+6)}{(x^3+2x)^2} \)
\( \implies \frac{dy}{dx} = \frac{2x^4+4x^2 - 3x^4-11x^2-6}{(x^3+2x)^2} \)
\( \implies \frac{dy}{dx} = \frac{-x^4-7x^2-6}{(x^3+2x)^2} \)
Now, we find the value of \( \frac{dy}{dx} \) at \( x = 1 \).
\( \left(\frac{dy}{dx}\right)_{x=1} = \frac{-(1)^4-7(1)^2-6}{((1)^3+2(1))^2} \)
\( \implies \left(\frac{dy}{dx}\right)_{x=1} = \frac{-1-7-6}{(1+2)^2} \)
\( \implies \left(\frac{dy}{dx}\right)_{x=1} = \frac{-14}{(3)^2} \)
\( \implies \left(\frac{dy}{dx}\right)_{x=1} = -\frac{14}{9} \)
In simple words: For each part, first find the derivative of the given fraction. In part (i), simplify the fraction before differentiating. Then, plug in the specific \(x\)-value into the derivative you found to get the numerical answer.

🎯 Exam Tip: Always look for opportunities to simplify the function before differentiating, especially if you can factor and cancel terms. This can make the differentiation process much easier. When applying the quotient rule, be careful with signs and polynomial expansion.

Question 20. Find the coordinates of the points on the curve \( y = \frac{x}{1-x^2} \) for which \( \frac{dy}{dx} = 1 \).
Answer:
Given the equation of the curve \( y = \frac{x}{1-x^2} \).
First, we differentiate \(y\) with respect to \(x\) using the quotient rule to find \( \frac{dy}{dx} \).
\( \frac{dy}{dx} = \frac{(1-x^2)\frac{d}{dx}(x) - x\frac{d}{dx}(1-x^2)}{(1-x^2)^2} \)
\( \implies \frac{dy}{dx} = \frac{(1-x^2)(1) - x(-2x)}{(1-x^2)^2} \)
\( \implies \frac{dy}{dx} = \frac{1-x^2 + 2x^2}{(1-x^2)^2} \)
\( \implies \frac{dy}{dx} = \frac{1+x^2}{(1-x^2)^2} \)
Now, we are given that \( \frac{dy}{dx} = 1 \). So, we set our derivative equal to 1.
\( \frac{1+x^2}{(1-x^2)^2} = 1 \)
\( \implies 1+x^2 = (1-x^2)^2 \)
\( \implies 1+x^2 = 1 - 2x^2 + x^4 \)
Rearrange the equation to solve for \(x\):
\( x^4 - 2x^2 - x^2 + 1 - 1 = 0 \)
\( \implies x^4 - 3x^2 = 0 \)
Factor out \(x^2\):
\( x^2(x^2 - 3) = 0 \)
This gives us two possibilities for \(x\):
\( x^2 = 0 \implies x = 0 \)
\( x^2 - 3 = 0 \implies x^2 = 3 \implies x = \pm \sqrt{3} \)
Now, we find the corresponding \(y\) values for these \(x\) values using the original equation \( y = \frac{x}{1-x^2} \).
When \( x = 0 \):
\( y = \frac{0}{1-(0)^2} = \frac{0}{1} = 0 \)
So, one point is \( (0, 0) \).
When \( x = \sqrt{3} \):
\( y = \frac{\sqrt{3}}{1-(\sqrt{3})^2} = \frac{\sqrt{3}}{1-3} = \frac{\sqrt{3}}{-2} = -\frac{\sqrt{3}}{2} \)
So, another point is \( \left(\sqrt{3}, -\frac{\sqrt{3}}{2}\right) \).
When \( x = -\sqrt{3} \):
\( y = \frac{-\sqrt{3}}{1-(-\sqrt{3})^2} = \frac{-\sqrt{3}}{1-3} = \frac{-\sqrt{3}}{-2} = \frac{\sqrt{3}}{2} \)
So, the third point is \( \left(-\sqrt{3}, \frac{\sqrt{3}}{2}\right) \).
Therefore, the coordinates of the required points on the curve are \( (0, 0) \), \( \left(\sqrt{3}, -\frac{\sqrt{3}}{2}\right) \), and \( \left(-\sqrt{3}, \frac{\sqrt{3}}{2}\right) \).
In simple words: First, find the derivative of the curve using the quotient rule. Then, set this derivative equal to 1 and solve for \(x\). Once you have the \(x\)-values, plug them back into the original curve equation to find their matching \(y\)-values. These \((x,y)\) pairs are the points you need.

🎯 Exam Tip: Remember that squaring both sides can introduce extraneous solutions, so it's vital to substitute the \(x\)-values back into the original condition (or the derivative equation) to ensure they are valid. Also, do not forget the \( \pm \) when taking square roots.